Question

3 rigid rods are joined to form an equilateral triangle. find the moment of inertia

Answer 1

Let the side be L then we know that moment of inertia at the centre of rod is ml^2/12  But we also need to account the perpendicular distance from its centroid to its side which is l/2√3  Thus I=I+I+I  =(IcM)+md²  =ML²/12+M(L/2√3)² =ML²/12+ML²/4×3 =ML²/12+ML²/12 =2ML²/12 =ML²/6 As there are 3 rods we multiply be 3  =3×ML²/6 =ML²/2